Problem: The lifespans of lions in a particular zoo are normally distributed. The average lion lives $10.8$ years; the standard deviation is $1.4$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a lion living less than $12.2$ years.
Solution: $10.8$ $9.4$ $12.2$ $8$ $13.6$ $6.6$ $15$ $68\%$ $16\%$ $16\%$ We know the lifespans are normally distributed with an average lifespan of $10.8$ years. We know the standard deviation is $1.4$ years, so one standard deviation below the mean is $9.4$ years and one standard deviation above the mean is $12.2$ years. Two standard deviations below the mean is $8$ years and two standard deviations above the mean is $13.6$ years. Three standard deviations below the mean is $6.6$ years and three standard deviations above the mean is $15$ years. We are interested in the probability of a lion living less than $12.2$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $68\%$ of the lions will have lifespans within 1 standard deviation of the average lifespan. The remaining $32\%$ of the lions will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({16\%})$ will live less than $9.4$ years and the other half $({16\%})$ will live longer than $12.2$ years. The probability of a particular lion living less than $12.2$ years is ${68\%} + {16\%}$, or $84\%$.